C= \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\) 

=\(\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\) 

= \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\) ( viet nguoc lai cho de nhin)

= \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\) 

= \(\frac{1}{100}-\left(1-\frac{1}{100}\right)\) 

= \(-\frac{49}{50}\)

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - .... - 1/3.2 - 1/2.1

\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+...+1-\frac{1}{2}\right)\)

\(C=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)=-\frac{1}{2}\)

thanh niên trên làm đúng

50 nếu ai thích sakura thì **** mình nếu ai thích sakura mà Ko **** mình thì chứng tỏ bạn Ko thích sakura

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\frac{99}{100}\)

\(C=\frac{-98}{100}=\frac{-49}{50}\)

Ủng hộ mk nha ^_-

 1/100-1/100.99-1/99.98-.....1/3.2-1/2.1

= 1/100-(1/100.99+1/99.98+.....+1/3.2+1/2.1)

=1/100-(1/1-1/2+1/2-1/3+...+1/98-1/99+1/99-1/100)

=1/100-(1/1-1/100)

=1/100-99/100

=-98/100

=-49/50

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{97.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\) 

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(\frac{1}{100}-C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\frac{1}{100}-C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{100}-C=1-\frac{1}{100}\)

\(C=C=\frac{1}{50}-1=-\frac{49}{50}\)

C=1/100-(1/100.99+1/99.98+...+1/3.2+1/2.1)

  =1/100-(1-1/2+1/2_1/3+...+1/99-1/100)

  =1/100-(1-1/100)

  =1/100-99/100

  =1/100 chọn cho mình nha!